Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 2x}{x - 8} = \dfrac{8x - 9}{x - 8}$
Explanation: Multiply both sides by $x - 8$ $ \dfrac{x^2 - 2x}{x - 8} (x - 8) = \dfrac{8x - 9}{x - 8} (x - 8)$ $ x^2 - 2x = 8x - 9$ Subtract $8x - 9$ from both sides: $ x^2 - 2x - (8x - 9) = 8x - 9 - (8x - 9)$ $ x^2 - 2x - 8x + 9 = 0$ $ x^2 - 10x + 9 = 0$ Factor the expression: $ (x - 1)(x - 9) = 0$ Therefore $x = 1$ or $x = 9$ The original expression is defined at $x = 1$ and $x = 9$, so there are no extraneous solutions.